Question

1. you know the height from which an object is dropped. which equation could you use to calculate the velocity as the object hits the ground?

a. ( v = h )
b. ( v = sqrt{2h} )
c. ( v = gh )
d. ( v = sqrt{2gh} )

Explanation:

Step1: Recall free - fall physics

In free - fall, the mechanical energy is conserved. The potential energy at height \(h\) (\(PE = mgh\)) is converted into kinetic energy (\(KE=\frac{1}{2}mv^{2}\)) just before hitting the ground (assuming initial velocity \(u = 0\)).

Step2: Equate potential and kinetic energy

Set \(PE=KE\), so \(mgh=\frac{1}{2}mv^{2}\).
We can cancel out the mass \(m\) from both sides of the equation (since \(m
eq0\)). We get \(gh=\frac{1}{2}v^{2}\).

Step3: Solve for velocity \(v\)

Multiply both sides of the equation \(gh=\frac{1}{2}v^{2}\) by 2: \(2gh = v^{2}\).
Then take the square root of both sides. Since velocity is a scalar quantity (we are considering the magnitude here) and \(h\) and \(g\) are positive, \(v=\sqrt{2gh}\).

Answer:

d. \(v = \sqrt{2gh}\)